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Question Number 199707 by SANOGO last updated on 08/Nov/23
study the convergence Σ_(n≥o) ^ sin(π(√(4n^2 +2 ))
$${study}\:{the}\:{convergence} \\ $$$$\underset{{n}\geqslant{o}} {\overset{} {\sum}}{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{2}\:\:\:\:}\right. \\ $$
Answered by witcher3 last updated on 09/Nov/23
(√(4n^2 +2))=2n(√(1+(1/(2n^2 )))) sin(2πn(√(1+(1/(2n^2 )))))=sin(2πn(√(1+(1/(2n^2 ))))−2πn) =sin(2πn((√(1+(1/(2n^2 ))))−1) (√(1+(1/(2n^2 ))))−1=(1/(2n^2 ((√(1+(1/(2n^2 ))))+1)))≤(1/(4n^2 )) sin(0)≤sin(2πn(√(1+(1/(2n^2 ))))−2πn)≤sin((π/(2n^2 )))≤sin((π/2)) positiv serie 2πn((√(1+(1/(2n^2 )))))−2πn =2πn(1+(1/(2n^2 ))−1+o((1/n^2 ))=(π/n)+o((1/n)) sin((π/n)+o((1/n)))∼(π/n);Σ(π/n) dv⇒Σsin(π(√(4n^2 +2))) dv
$$\sqrt{\mathrm{4n}^{\mathrm{2}} +\mathrm{2}}=\mathrm{2n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }} \\ $$$$\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}\right)=\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}−\mathrm{2}\pi\mathrm{n}\right) \\ $$$$=\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}−\mathrm{1}\right)\right. \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}+\mathrm{1}\right)}\leqslant\frac{\mathrm{1}}{\mathrm{4n}^{\mathrm{2}} } \\ $$$$\mathrm{sin}\left(\mathrm{0}\right)\leqslant\mathrm{sin}\left(\mathrm{2}\pi\mathrm{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}−\mathrm{2}\pi\mathrm{n}\right)\leqslant\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}^{\mathrm{2}} }\right)\leqslant\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{positiv}\:\mathrm{serie} \\ $$$$\mathrm{2}\pi\mathrm{n}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }}\right)−\mathrm{2}\pi\mathrm{n} \\ $$$$=\mathrm{2}\pi\mathrm{n}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}^{\mathrm{2}} }−\mathrm{1}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)=\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right. \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{n}}+\mathrm{o}\left(\frac{\mathrm{1}}{\mathrm{n}}\right)\right)\sim\frac{\pi}{\mathrm{n}};\Sigma\frac{\pi}{\mathrm{n}}\:\:\mathrm{dv}\Rightarrow\Sigma\mathrm{sin}\left(\pi\sqrt{\mathrm{4n}^{\mathrm{2}} +\mathrm{2}}\right)\:\mathrm{dv} \\ $$

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